Let $g (x) = x^{3} - 6 x^{2} + 9 x$ on the closed interval $[0, 3]$ .

hard 4 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c = 1

and

g^{''} (1) = - 6

The correct answer is $c = 1$ and $g^{''} (1) = - 6$ — Thus the required interior point is $c = 1$ and $g^{''} (1) = - 6$ .

Step-by-step solution

Check continuity on $[0, 3]$ : $g$ is a polynomial, hence continuous everywhere.
Check differentiability on $(0, 3)$ : polynomials are differentiable everywhere, so differentiable on the open interval.
Evaluate endpoints: $g (0) = 0^{3} - 6 \cdot 0^{2} + 9 \cdot 0 = 0$ , $g (3) = 27 - 54 + 27 = 0$ . Since $g (0) = g (3)$ , the equal‑value condition holds.
All three hypotheses of Rolle’s theorem are satisfied.
Compute the first derivative: $g^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ .
Set $g^{'} (c) = 0$ and look for solutions inside $(0, 3)$ : $c = 1$ and $c = 3$ . The point $c = 3$ is an endpoint, so the only interior solution is $c = 1$ .
Compute the second derivative: $g^{''} (x) = 6 x - 12$ .
Evaluate at $c = 1$ : $g^{''} (1) = 6 (1) - 12 = - 6$ .
Thus the required interior point is $c = 1$ and $g^{''} (1) = - 6$ .

Common mistake: Including the endpoint

x = 3

as a solution of

g^{'} (c) = 0

for the interior requirement.

Concept tested: Rolle’s theorem and critical points